# Gibbs Free Energy Calculation

## Main Question or Discussion Point

ΔG=ΔH-TΔSsystem and ΔSsystem=qp/T and isn't qp=ΔH? Which means ΔG=0 always which is obviously wrong.

But the derivation for Gibbs Free Energy is based on qsurrounding=-qsystem so shouldn't the term be equals to 0?

Ygggdrasil
Gold Member
2019 Award
ΔG=ΔH-TΔSsystem and ΔSsystem=qp/T and isn't qp=ΔH? Which means ΔG=0 always which is obviously wrong.
ΔSsystem=qp/T is true only for reversible processes. What do you know about the ΔG of a reversible process?

ΔSsystem=qp/T is true only for reversible processes. What do you know about the ΔG of a reversible process?
It is 0. So for a reversible process qp=ΔH and so ΔG=0?

For an irreversible process how would the ΔS look like?

Thanks

Chestermiller
Mentor
It is 0. So for a reversible process qp=ΔH and so ΔG=0?

For an irreversible process how would the ΔS look like?

Thanks
Are you aware that ΔG and ΔS are functions only of the initial and final thermodynamic equilibrium states of a system, and are independent of path between the two states, whether reversible or irreversible? From what you have learned, do you know how to determine the change in S between two thermodynamic equilibrium states of a system? Are you familiar with the equation dG = -SdT + VdP, which applies to a differential change in temperature and pressure between two closely neighboring thermodynamic equilibrium states of a system?

I think your real question is "how do one determine ΔG and ΔS between two thermodynamic equilibrium states of a system."

Chet

It is possible for ΔSsystem ≠ 0, even when qP = 0. Therefore, qP = 0, does not necessarily lead to ΔG = 0.